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A Visualization Tool for the Vibration of Euler-Bernoulli and Timoshenko Beams

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A Visualization Tool for the Vibration of Euler-Bernoulli and Timoshenko Beams The Mathematica Journal ® A Visualization Tool for the Vibration of Euler-Bernoulli and Timoshenko Beams Prithvi Akella , Evan G. Hemingway, Oliver M. O’Reilly This article details the background material on the classic Euler- Bernoulli and Timoshenko beam theories. A tool designed for visualizing the deformation of beams in static equilibrium and dynamic vibration under an applied loading is also demonstrated. The treatment of these topics is intended to be accessible to upper-division undergraduate engineering students. The Euler-Bernoulli Beam ■ Mathematica_Paper_4.nb 5/5/17 The Mathematica Journal volume:issue year Wolfram Media, Inc. © 2 Author(s) Figure 1. A deformed Euler-Bernoulli beam and its reference configuration in which the centerline is straight and parallel to the x-axis. The cross-sections of the beam remain plane and orthogonal to the centerline after deformation. The upward displacement of a material point on the centerline at a location x in the reference configuration is denoted by w(x,t) and the counterclockwise rotation of a cross-section at a location x in the reference configuration is denoted by the angle (x,t). θ The Euler-Bernoulli beam model is the simplest linear beam theory and is dis- cussed in most elementary textbooks on vibration (see, e.g. [6]). It works particularly well for slender beams or beams with a high shear modulus, where deformations due to bend- ing are more significant than those due to shear. The centerline of the beam is defined to be the line of geometric centers along the length and it is parametrized by the coordinate x in the reference configuration. We assume the centerline to be inextensible while the cross-sections remain plane and normal to the centerline. To apply the theory, we require knowledge of the following beam properties: , the volume mass density, A, the cross-sec- tional area, , the length, E, the modulus of elasticity,ρ and I, the area moment of inertia. Re- ferring to Figureℓ 1, if w and u are the respective vertical and horizontal displacements of a point on the centerline relative to a straight reference configuration, then geometric consid- erations show that w u sin , 1 cos , 1 ∂ x ∂ x = (θ) + = (θ) ( ) where∂ the coordinate x parametrizes∂ the initially straight centerline. Assuming small deflec- tions, these relations simplify to The Mathematica Journal volume:issue year Wolfram Media, Inc. Mathematica_Paper_4.nb 5/5/17 © Article Title 3 w u , 0. 2 ∂ x ∂ x ≃ θ ≃ ( ) The ∂signed curvature∂ of the centerline is related to the angle as: κ θ . 3 ∂θx κ = ( ) 2 Thus, for ∂small deflections of the beam, can be approximated by w . ∂ x2 κ ∂ Figure 2. TRY TO MAKE NON-FUZZY A differential element of an Euler-Bernoulli beam at position x = x and time t = t . The bending moment and shear force at x +dx are * * * approximated using Taylor series expansions: M x dx, t ) = M x , t ) + M x , t dx * * * * ∂ x * * and V x dx, t ) = V x , t ) + V x , t dx. ( + ( ∂ ( ) * * * * ∂ x * * ( + ( ∂ ( ) Referring to Figure 2, the forces acting on a cross-section of the beam include a shear force V and an axial force, which are the resultants of shear and normal tractions inte- grated over the cross-sectional area. The axial force ensures that the centerline remains in- extensible. The curvature of the beam is due to a bending moment, M, which is the resul- tant torque in the z-direction due to normal tractions about the centerline. We assume the classic constitutive relation 2 w M EI , 4 ∂ x2 = ( ) ∂ Mathematica_Paper_4.nb 5/5/17 The Mathematica Journal volume:issue year Wolfram Media, Inc. © 4 Author(s) 2 where we have substituted w for the centerline curvature and EI is known as the flexural ∂ x2 rigidity. ∂ In Euler-Bernoulli beam theory, the inertia of the rotating differential elements and the deformation due to shear is assumed to be negligible, so that cross-sections remain nor- mal to the centerline. Consequently, the balance of angular momentum for an element of the beam yields the relation M 0 m V. 5 ∂ x = + + ( ) Here, m =∂ m(x,t) is an applied moment per unit length about the centerline of the beam in the -y-direction. The balance of linear momentum in the z-direction yields 2 w V A f . 6 ∂ t2 ∂ x ρ = + ( ) Here, A∂ is the linear∂ mass density of the beam, and f = f(x,t) is an applied force per unit length ρin the z-direction. Together, m and f make up the applied loading. Assuming no ap- plied forces in the horizontal direction, small deformations, and inextensibility, the bal- ance of linear momentum in the horizontal direction will show that the axial force in the beam is constant. Combining equations (4), (5), and (6) and assuming that A and EI are indepen- dent of x (i.e. the beam is homogenous and prismatic) yields a constantρ coefficient fourth order linear PDE which is the equation of motion for the Euler-Bernoulli beam: 4 w 2 w fa EI A , 7 ∂ x4 ∂ t2 - = ρ ( ) ∂ m ∂ where fa = f - is defined as the total applied loading. To solve this partial differential ∂ x equation, we require∂ the initial conditions w w x, t , x, t , 8 0 ∂ t 0 ( ) ( ) ( ) at time t = t0 along∂ with the four independent boundary conditions featuring w 2 w 3 w w x , t , x , t , x , t , x , t , 9 * ∂ x * ∂ x2 * ∂ x3 * ( ) ( ) ( ) ( ) ( ) at locations x∂ = x . For instance,∂ x is∂ typically chosen to be 0 or . To obtain solutions * * 2 for the static case, we first set w 0 in equation (7). Then, one performsℓ four simple in- ∂ t2 tegrations while applying four∂ independent= boundary conditions to obtain the solution. The resulting displacement w = w(x) is the solution up to a rigid translation and constant velocity. The following function defines the deformed centerline and generates the visual- ization in Figure 3. The Mathematica Journal volume:issue year Wolfram Media, Inc. Mathematica_Paper_4.nb 5/5/17 © Article Title 5 StaticEulercenterline Emod , Iarea , TotalLength , Lsupport , Rsupport[ , t _, b : _Module , _ Clear position,_ w, q,_ eqn1,_ _ eqn2,] = eqn3,[ eqn4,{} eqn5, solution[ ; q t b; ] eqn1= + w'''' x q Emod Iarea ; eqn2 = Switch[ Lsupport,] == /( Fix, w 0) 0, S, w 0 0, Free,= w'' [0 0 ; [ ] [ ] eqn3 Switch[ ]Lsupport, ] Fix, w' 0 0, S, w'' 0 0, Free,= w'''[0 0 ; [ ] [ ] eqn4 Switch[Rsupport,] ] Fix, w TotalLength 0, S,= w TotalLength[ 0, Free,[ w'' TotalLength] 0 ; eqn5 [Switch Rsupport,] Fix, w' TotalLength[ 0,] ] S,= w'' TotalLength[ 0, Free,[ ] w''' TotalLength[ ]0 ; solution[ DSolve ]eqn1, ] eqn2, eqn3, eqn4, eqn5 , w x , x,= 0, TotalLength[{ ; } position[ ] { x : Evaluate solution}] 1 1 2 ; Return position[ _] = ; [ [[ ]][[ ]][[ ]]] [ ]] StaticEulerSections position , TLength : Module , Clear newposition,[ slope,_ shear, moment,_] = radius,[{} theta,[ crosssection, graphing ; newposition x : position x ;] slope x : [newposition_] = ' x[;] shear[x_] := position ''' [x ]; moment[ _]x =: position ''[x]; radius[x_] := 1 newposition[ ] '' x 5; theta [x _]: =ArcTan ( / newposition[ ' ])x /; crosssection[ _] = var1 ,[ var2 , var3 [, constant]] , TotalLength[ , newposition_ _ , theta_ : _ If var1 var2_ && var2 var3,_ _] = Graphics[ ≥ ≥ Line [ var2[ Sin constant theta var2 TotalLength 10, {{constant+ newposition[ var2[ ]] / Cos constant theta [var2 ]-TotalLength 10 , var2 [ Sin constant theta[ ]]var2 TotalLength/ } 10, {constant- newposition[ var2[ ]] / Cos constant theta [var2 ]+TotalLength 10 , ; [ [ ]] / }}]] graphing{}] var1 , n , place , constant , TotalLength , newposition[ _ , theta_ :_ _ _ _ _] = Mathematica_Paper_4.nb 5/5/17 The Mathematica Journal volume:issue year Wolfram Media, Inc. © 6 Author(s) Module sep, sections , sections 0 Range n 1 ; sep [TotalLength{ n;} = [ + ] For =i 1, i n /1, i , sections[ = i<= +crosssection++ var1, i 1 sep, place,[[ constant,]] = TotalLength,[ newposition,( - ) theta ; Return sections ; ]] Return Manipulate[ ]]Show Plot[ d newposition[ x[ , x, l, u 0.01 , PlotRange[ 0, TLength[ ] { , c,+ c ,} AxesLabel {{"x", "w x "} ,{- }} If line, Graphics { [ ] }] [Line u Sin b[ theta u TLength 10, b[{{ newposition+ [ u [Cos]] b theta u/ TLength 10 , u Sin b theta[ u]- TLength[ 10,[ ]] / } {b- newposition[ [u ]]Cos b theta/ u TLength 10 , Plot 0, x, 0, 1 [ ,]+ [ [ ]] / }}]] If line,[ graphing{ }]]u, number, l, b, TLength, [newposition, theta[ , Plot 0, x, 0, 1 , u, 0.01, TLength 0.01] , [ { }]]] {Item "u sets RHS- x axis} graph limit", Alignment[ - Left ,- l, 0.01, TLength ] 0.01 , {Item "l sets the- LHS x}axis graph limit", Alignment[ Left , - b, 10000 , ] {Item } "b[ magnifies the graph and all associated material cross sections, angles, etc ", Alignment (Left ,- ) c, TLength, ] {Item "c sets} the pos neg values on the y axis", Alignment[ Left , / - number, 5 , ] {Item } "number[ sets the number of cross sections the program will animate", Alignment- Left , line, True, False , ] {Item "Line{ switches}} the cross sections on off", Alignment[ Left ; - / ]]]] The Mathematica Journal volume:issue year Wolfram Media, Inc. Mathematica_Paper_4.nb 5/5/17 © Article Title 7 w x [ ] 4 2 0 x 1 2 3 4 5 2 - 4 - Figure 3. Example of a static deformation of an Euler-Bernoulli beam under a sinusoidal applied traction load. Given pin joints as support structures, the four independent bound- ary conditions require that V = 0 and M = 0 at both ends of the beam. Here, we used the following data: Emod = 100 Pa Iarea: 5 m4 TLength: 5 m t = Sin[ x] b = 0 π Free Vibrations □ The more interesting case concerns the dynamic vibrations of the beam. To proceed, we first set f = 0 and m = 0 and solve the free vibration problem. The governing equation (7) is written in standard form as 4 w 2 w c2 0, 10 ∂ x4 ∂ t2 + = ( ) where ∂ ∂ EI c . 11 A = ( ) Next, we assumeρ that the mode shape profile is independent of time so that the desired so- lution is separable: w x, t W x Q t .
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